Still stuck designing a diamond pattern leaded window

[Rob Edwards]

Hello again Math Wizards.


For context this is a subsequent problem I am facing after the first problem was solved here

Stuck designing a diamond pattern leaded window.

Hello everyone. This is my first post - apologies if I don't follow correct forum etiquette. I am a traditional woodworker designing a window in CAD. My maths is very rusty (30 years since I did A-Levels in the UK) I attach a picture of the problem I'm trying to solve and repeat the...

www.freemathhelp.com

I am attempting to work out how many diamonds I need of a given proportion of no greater than a maximum width to fill a designated total width.
The diamonds have a border of known size.

Here's an image that hopefully paints a clearer picture n is the number of panels of width a


So far I have reduced the problem to





From here

I know that the proportion is equal to the tangent and I know f (it is half the border width) so if I can solve this I think I can manage the rest.

With the following as a starting point I've tried all the obvious things in my limited repertoire.

p = d / e = tan(a)

g^2 = e^2 + d^2

d = f / cos(a)

e = f / sin(a)

sin(a) = d / g

cos(a) = e / g

It's way above my pay grade.. Any help would be greatly appreciated

 

[Rob Edwards]

oooh! Have I got it? feels like I've made a daft mistake

 

[Dr.Peterson]

All that is correct; but if you simplify your result you find that it is

[MATH]e = \sqrt{f^2\left(\frac{1}{\tan^2\alpha}+1\right)} = f\sqrt{\frac{\cos^2\alpha}{\sin^2\alpha}+1} = f\sqrt{\frac{\cos^2\alpha+\sin^2\alpha}{\sin^2\alpha}} = \frac{f}{\sin\alpha}[/MATH]​

You already had that at the top right of your paper! What you've really done amounts to reversing what I just did.

And wanted this form (with tan) because you wanted to use p. So your final result is [MATH]e = f\sqrt{\frac{1}{p^2}+1}[/MATH], right?

That can be simplified a little more, to [MATH]e =\frac{f}{p}\sqrt{p^2+1}[/MATH]
But ultimately, good work! You really did more than you thought you could.

 

[Rob Edwards]

Thankyou again Dr Peterson.

I think my initial difficulty was because I substituted in p too soon. After I posted I looked on wikipedia and found the ways of expressing one trig function in terms of the others. I didn't see then but I see now I could have gone a more direct route.

Your simplification will really help the readability of my code.

I can't believe how much pleasure it gives me to solve these problems, I wish my younger self was as interested as I am now.

ps I will endeavour to learn the syntax you use for presenting the formulas so I don't have to take photos. It will also help immensely with my documentation.

I'm sure I'll be back soon enough. Adios